I am asked to find the total number of ways I can distribute 12 distinguishable objects amongst 3 distinguishable groups (1,2,3) such that 1 receives at least 2 objects and 2,3 receive at least one each. I sense this is related to PIE but I am struggling in applying the principle to this case. This is my attempt so far:
Total number of outcomes including unusable ones = $3^{12}$ = 531441
Let |X1| be the outcomes where x1 received no objects and let |X2| be the set of outcomes where x2 received no objects and let |X3| be the set of outcomes where x3 received no objects. There are $2^{12}$ possible outcomes here.
Then |X1$\cap$X2|, or |X1$\cap$X3| = $1^{12}$ and |X1$\cap$X2$\cap$X3| = $0^{12}$
Let |X11| be the set of outcomes where 1 received only 1 object = $2^{11}$
Now to exclude outcomes where any group received 0 objects:
${3 \choose 1}$$2^{12}$ - ${3 \choose 2}$$1^{12}$ + ${3 \choose 3}$*$0^{12}$ = 12286
531441 - 12286 = 519155
Now to exclude outcomes where any x1 received just one object:
519155 - $2^{11}$ = 517107.
Am I close? All advice and constructive criticism would be greatly appreciated!
You have the correct method for computing the number of distributions where no one receives zero objects, though your number is off by one because of an arithmetic error: $$ 3\cdot 2^{12}-3\cdot 1^{12}+1\cdot 0^{12}=1228{\color{red}5}. $$ However, the number of ways where the first person gets exactly one object is not $2^{11}$. The $2^{11}$ counts the number of ways to distribute $11$ objects to the second and third persons, but you have not factored in the number of ways to choose which object person number one receives.
If you fix these two issues, your solution will be totally correct.